Physics · Thermodynamics

JEE Advanced 2018 — Paper 2 — Question 11

One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant $\mathrm{R}=8.0 \mathrm{~J}$ $\mathrm{mol}^{-1} \mathrm{~K}^{-1}$ , the decrease in its internal energy, in Joule, is $\_\_\_\_$ .

Answer: 900

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}}=\left(\frac{\mathrm{V}_{\mathrm{i}}}{\mathrm{V}_{\mathrm{f}}}\right)^{\gamma-1}=\left(\frac{1}{8}\right)^{2 / 3}$

$\mathrm{T}_{\mathrm{f}}=\frac{\mathrm{T}_{\mathrm{i}}}{4}=\frac{100}{4}=25 \mathrm{~K}$

$\therefore \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{V}}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right)$

$=1 \times \frac{3 R}{2}(25-100)=\frac{3}{2} \times 8 \times(-75)=-900$ Joule

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Physics
Chapter: Thermodynamics
Topic: Different Thermodynamic Processes

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PVdiagram below. Among these four…

Answer: 900

Step-by-step solution

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